HMMT February

Solution:

Let $H_a$ be the foot of the altitude from $A$ to $BC$. Since $AE$ bisects $\angle H_a AV$, by the angle bisector theorem $\frac{AH_a}{H_aE}=\frac{AV}{VE}$. Note that $\triangle AH_aE \sim \triangle ANA'$ are similar right triangles, so $\frac{AN}{NA'}=\frac{AH_a}{H_aE}$.

Let $R$ be the radius of $\Gamma$. We know that $AA'=2R$, so $AN=\sqrt{AA'^2-NA'^2}=\sqrt{4R^2-81}$ and $AV=AA'-VA'=2R-7$. Therefore

$$\frac{\sqrt{4R^2-81}}{9}=\frac{AN}{NA'}=\frac{AH_a}{H_aE}=\frac{AV}{VE}=\frac{2R-7}{6}$$

The resulting quadratic equation is

$$0=9(2R-7)^2-4\left(4R^2-81\right)=20R^2-252R+765=(2R-15)(10R-51)$$

We are given that $ABC$ is acute so $VA'<R$. Therefore $R=\frac{15}{2}$.

Solution:

Let $\Psi$ denote inversion about $A$ with radius $\sqrt{AB \cdot AC}$ composed with reflection about $AE$. Note that $\Psi$ swaps the pairs $\{B, C\},\{E, N\}$, and $\{H_a, A'\}$. Let $K=\Psi(V)$, which is also the second intersection of $AH_a$ with $\Gamma$. Since $AE$ bisects $\angle KAA'$, we have $NK=NA'=9$. By the inversion distance formula,

$$NK=\frac{AB \cdot AC \cdot VE}{AE \cdot AV}=\frac{AE \cdot AN \cdot VE}{AE \cdot AV}=\frac{AN \cdot VE}{AV}$$

This leads to the same equation as the previous solution.

We are given that $ABC$ is acute so $VA'<R$. Therefore $R=\frac{15}{2}$.